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vovangra [49]
3 years ago
14

A tour company has a boat that sits at most 40 passengers. For a cruise on the Hudson River they charge $12 per adult and $5 per

child. They will only take the boat out if they have sold at least $250 worth of tickets. Let a be the number of adults who take the cruise and c be the number of children. Will they take the boat out if they 14 adults and 9 children buy tickets? Justify.
Mathematics
1 answer:
EleoNora [17]3 years ago
5 0

Answer:

No

Step-by-step explanation:

For the adults:

$12 times the 14 adults which is: $168

Then for the kids:

$5 times the 9 kids which is: $45

Add next:

$168 plus the $45 equals $213

A(14)+b(5)=c

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Answer:

x=4x therefore the answer is 4x

Step-by-step explanation:

x^2-8x+16=0

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3 years ago
Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi
BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

\mu = 6*0.6*3 + 6*0.4*1 = 13.2

The standard deviation is:

\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

\mu = 6*0.6 = 3.6

The probability of scoring four goals is:

P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122

Probability of two assists:

40% of 2 are assists, which means that:

\mu = 6*0.4 = 2.4

The probability of getting two assists is:

P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127

Probability of four goals and two assists:

P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05

0.05 = 5% probability that he has four goals and two assists in one game

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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